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2 years ago

What are metals in chemistry

Chemistry

2 answers:

elena-14-01-66 [18.8K]2 years ago

6 0

usually referring to chemical elements that are solid

Masja [62]2 years ago

4 0

pononium ,and silver

Explanation:

hope it helped

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Dalam ruang 1 liter, gas metana (ch4) direaksikan dengan 6 mol uap air sesuai reaksi:ch4(

kakasveta [241]

In the space of 1 liter, methane gas (ch4) is reacted with 6 moles of water vapor according to the reaction: if at equilibrium state is obtained 4 moles of hydrogen gas, how many moles of methane gas is needed for the equilibrium reaction?

the reaction is

CH4(g) + 2H2O(g) ----> CO2(g) + 4H2 (g)

Kc = 16 / 3

Kc = [CO2] [H2]^4 / [CH4] [H2O]^2

given :

equilibrium concentration

[H2] = 4 moles

so equilibrium concentration of CO2 must be 1 mole

equilibrium concentration of H2O = 6 - 2 = 4

putting values

16 /3 = [1] [4]^4 / [CH4] [4]^2

[CH4] = 0.333 moles

so moles of CH4 required = 1.33 moles

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3 years ago

How many formula units make up 25.8g of magnesium chloride (MgCl2)? Express the number of formula units numerically.

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2 years ago

Guys i need help What is the chemical equation for foam fight

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Answer:

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A 50.0 mL solution of 0.141 M KOH is titrated with 0.282 M HCl . Calculate the pH of the solution after the addition of each of

Kobotan [32]

Answer:

pH =1 2.84

Explanation:

First we have to start with the reaction between HCl and KOH:

$HCl~+~KOH->~H_2O~+~KCl$

Now for example, we can use a volume of 10 mL of HCl. So, we can calculate the moles using the molarity equation:

$M=\frac{mol}{L}$

We know that $10mL=0.01L$ and we have the concentration of the HCl $0.282M$ , when we plug the values into the equation we got:

$0.282M=\frac{mol}{0.01L}$

$mol=0.282*0.01$

$mol=0.00282$

We can do the same for the KOH values ( $50mL=0.05L$ and $0.141M$ ).

$0.141M=\frac{mol}{0.05L}$

$mol=0.141*0.05$

$mol=0.00705$

So, we have so far 0.00282 mol of HCl and 0.00705 mol of KOH. If we check the reaction we have a molar ratio 1:1, therefore if we have 0.00282 mol of HCl we will need 0.00282 mol of KOH, so we will have an excess of KOH. This excess can be calculated if we substract the amount of moles:

$0.00705-0.00282=0.00423mol~of~KOH$

Now, if we want to calculate the pH value we will need a concentration, in this case KOH is in excess, so we have to calculate the concentration of KOH. For this, we already have the moles of KOH that remains left, now we need the total volume:

$Total~volume=50mL+10mL=60mL$